# Calulate area when two circle intersect each other [duplicate]

I have Sphere with radius R. Inside the sphere I have Two circle. One circle is fixed defined by $$\alpha$$. Another circle can rotate and the orientation of that circle can be defined by $$\beta$$. From the centre of the sphere until the edge of the movable circle create $$\alpha$$ angle with $$z$$ axis. $$h$$ and $$a$$ are distance from the origin and base radius of this circle .

Now when both circle intersect each other what will be the surface area of intersection? Using spherical coordinate system $$r,\theta,\phi$$, I know that my $$\theta$$ limit goes from $$\gamma$$ to $$\alpha$$. But I am not getting the limit for $$\phi$$ with this given parameters. I am trying to calculate the intersected surface area using only calculus and trigonometry for relevancy of my future problem.

## marked as duplicate by David K, YuiTo Cheng, Daniele Tampieri, José Carlos Santos, Ak19Jul 26 at 12:57

• @YuriyS I have edited the picture. Is it clear now? – T. an Jul 25 at 8:18
• To the edited version: you mean the area of a circular segment on the horizontal circle which is cut off by the inclined circle? – Yuriy S Jul 25 at 8:20
• Or the part of the sphere surface cut off by the two circles? This seems more likely – Yuriy S Jul 25 at 8:21
• @YuriyS It is the whole surface area in between two circle's intersection. – T. an Jul 25 at 8:21
• Yeah part of the sphere cut off by two circle. – T. an Jul 25 at 8:22

I'll change the problem statement a little so I can use Cartesian coordinates at first.

Also let $$R=1$$, since we can always multiply the answer by $$R^2$$.

Let Circle 1 lie in the horizontal plane with elevation $$d$$ (Plane 1).

Let Circle 2 lie in the plane defined by the radius vector $$\vec{h}$$ (Plane 2).

Equation of Plane 1:

$$z=d$$

Equation of Plane 2:

$$h_x x+h_y y+h_z z=h^2$$

Equation of the sphere:

$$x^2+y^2+z^2=1$$

Now we transition to spherical coordinates:

$$x = r \sin \theta \cos \phi$$

$$y = r \sin \theta \sin \phi$$

$$z = r \cos \theta$$

Which gives us:

$$r=1$$

$$\cos \theta=d$$

$$h_x \sin \theta \cos \phi+h_y \sin \theta \sin \phi+h_z \cos \theta=h^2$$

We now have equations to determine the intersection points of the two circles on the sphere.

$$\pm \sqrt{1-d^2}(h_x \cos \phi+h_y \sin \phi)+h_z d=h^2$$

$$h_x \cos \phi+h_y \sin \phi= \pm\frac{h^2-h_z d}{\sqrt{1-d^2}}$$

Define:

$$h_x=\cos \delta, \qquad h_y= \sin \delta$$

Then we have:

$$\cos (\phi-\delta)=\pm\frac{h^2-h_z d}{\sqrt{1-d^2}}$$

$$\phi= \delta+\arccos \left(\pm \frac{h^2-h_z d}{\sqrt{1-d^2}} \right)$$

$$\phi= \arccos h_x+\arccos\left(\pm \frac{h^2-h_z d}{\sqrt{1-d^2}} \right)$$

$$\theta=\arccos d$$

We found the intersection point coordinates (note also the condition for their existence):

$$-1 \leq \frac{h^2-h_z d}{\sqrt{1-d^2}} \leq 1$$

I think now half of the problem is solved, since we can use those points to set up the limits of integration on the surface of the sphere.

• Thanks for your great try...So for the $\phi$ another limit suppose to be $\pi-\phi$ right? – T. an Jul 25 at 9:59
• @T.an, not exactly, see my edit. I forgot that I have two solutions for $\sin \theta$ – Yuriy S Jul 25 at 10:04
• so one is for when circle goes up from right side and crossing each other and another is for left side when it makes reverse crossing? – T. an Jul 25 at 10:07
• @T.an, you make my head hurt with this question :). I'm better at algebra than geometry, so I hope you figure it out – Yuriy S Jul 25 at 10:08
• @T.an, important: my $\gamma$ is not the same as yours. I will edit – Yuriy S Jul 25 at 10:36